Advanced Calculus, part I: Analysis and Topology in Normed Spaces

You do this one.

Label the endpoints of each closed interval so that

Observe that the nested condition means that

for all \(k\text{.}\) So the set \(A=\left\{\alpha_k\middle\vert k\in\mathbb{N}\right\}\) is bounded above and \(B=\left\{\beta_k\middle\vert k\in\mathbb{N}\right\}\) is bounded below.

Set \(\alpha=\sup A\) and \(\beta=\inf B\text{.}\) The claim is any \(p\in\left[\alpha,\beta\right]\) is in the intersection \(\bigcap_{n\in\mathbb{N}} I_n\text{.}\)

To see this, notice that if \(p\in[\alpha,\beta]\text{,}\) then \(p\geq \alpha\text{.}\) Since \(\alpha\) is an upper bound for \(A\text{,}\) this means \(p\geq \alpha_k\) for all \(k\text{.}\) Similarly, \(p\leq\beta\leq\beta_k\text{.}\) So \(\alpha_k\leq p\leq\beta_k\text{,}\) that is, \(p\in I_k\text{.}\)

We need to show that \([\alpha,\beta]\) is nonempty, which is equivalent to \(\alpha\leq\beta\text{.}\) Take any \(\beta_k\in B\) and any \(\alpha_n\in A\text{.}\) If \(n\geq k\text{,}\) then

on the other hand if \(n\leq k\text{,}\) then

In either case, \(\alpha_n\leq \beta_k\text{.}\) So each element \(\alpha_n\in A\) is a lower bound for \(B\text{,}\) hence is less than or equal to \(\beta=\inf B\) (because \(\inf B\) is the greatest lower bound). Therefore, \(\alpha_n\leq \beta\) for all \(n\text{.}\) But this makes \(\beta\) an upper bound for \(A\text{.}\) Since \(\alpha=\sup A\) is the least upper bound, this means \(\alpha\leq \beta\text{.}\)

Suppose \(\mathbb{N}\) were bounded. Then there would be \(B=\sup \mathbb{N}\text{.}\) Using \(\epsilon=1\) in the \(\epsilon\)-characterization, there is some natural number \(M\) with \(B-1\lt M\lt B\text{.}\)

If we add 1 across the board, we get

But \(M+1\) is a natural number, so \(B\) isn't an upper bound after all.

This contradiction establishes the claim.

First consider the case that \(y\leq 0\text{.}\) Then \(n=1\) works. So from now on, we can assume \(y\gt 0\text{.}\)

If the proposition were false, there would be \(x,y\) so that \(nx\leq y\) for all \(n\in\mathbb{N}\text{.}\) In other words, \(A=\left\{nx\middle\vert n\in\mathbb{N}\right\}\) is bounded above. By completeness, this set has a least upper bound, which we'll call \(B\text{.}\)

Consider \(B-x\text{.}\) Since \(x\gt 0\text{,}\) \(B-x\lt B\text{.}\) So there is \(N\in\mathbb{N}\) so that

which we can rearrange to

But \((N+1)x\in A\text{,}\) which means that \(B\) wasn't an upper bound to begin with.

This contradiction establishes the proposition.

Suppose not. That is, there would be some special \(\epsilon\gt 0\) so that \(\frac{1}{n}\geq \epsilon\) for all natural numbers \(n\text{.}\) But then \(\frac{1}{\epsilon}\geq n\) for all \(n\in \mathbb{N}\text{,}\) i.e. \(\frac{1}{\epsilon}\) would be an upper bound for the natural numbers.

This will be a class activity.

If \(x\lt 0\lt y\text{,}\) then we use \(r=0\text{.}\)

If \(x=0\lt y\text{,}\) then Proposition 1.3.12 gives a rational of the form \(\frac{1}{n}\text{.}\)

If \(x\lt y=0\text{,}\) then applying Proposition 1.3.12 gives \(\frac{1}{n}\) with \(0\lt \frac{1}{n}\lt -x\text{,}\) so \(x\lt -\frac{1}{n}\lt 0\text{.}\)

Consider the case \(0\lt x\lt y\text{.}\) Then \(y-x\gt 0\text{,}\) so Proposition 1.3.12 gives \(n\in\mathbb{N}\) with \(\frac{1}{n}\lt y-x\text{,}\) or equivalently,

which we'll rearrange as

Now because \(\mathbb{N}\) is not bounded, there is some natural number \(m\) with \(nx\lt m\text{.}\) In fact, let's choose the smallest such natural number. Because \(nx\) and \(ny\) are at least 1 apart, we must have \(m\lt ny\text{.}\)

Thus we have

and dividing through, we get

as desired.

The remaining case is what happens with \(x\lt y\lt 0\text{.}\) But then \(0\lt -y\lt -x\text{,}\) so by the case we just handled, there is \(q\in\mathbb{Q}\) with \(-y\lt q\lt -x\text{.}\) Use \(r=-q\text{.}\)